As you will see in Experiment 9, a string connecting a cart and a hanger goes ov

Question

As you will see in Experiment 9, a string connecting a cart and a hanger goes over a pulley, as in the sketch below on the left. The pulley has tensions T1 and T2 acting upon it, as shown below in the center. Write an expression for the total torque vector on the pulley in terms of the shown variables. Assume no slip between the string and the pulley throughout this problem.

Given that the pulley has a moment of inertia of I = 1.00 x 10-4 kg m2, T1 = 4.40 N, T2 = 4.39 N, and r

= 2.50 cm, what is the magnitude of the angular acceleration of the pulley?

What is the direction of the angular acceleration of the pulley (i.e. unit vector)?

The pulley starts rotating at t = 0 s. At t = 10.0 s, through how many radians has it revolved?

What is the angular speed at t = 10.0 s?

Explanation / Answer

first we have to find out resultant force by the help of paralelogram law of vectors

total torque=sqrt(T1^2+T2^2+2T1T2cos45*)

                   =sqrt(4.4^2+4.39^2+2 x 4.4 x4.39 x c0s 45)

                =65.95N

     T1r-T2r=Ialpha

       T1r-T2r=I xa/r

      a= (T1-T2) x r^2/I=0.01 x 6.25 x10^-4/1 x 10^-4

        =0.0625m/s^2

angular acceleration = a/r

                           =0.0625/0.025

                            =2.5 rad/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.